Electronic Devices

Syllabus

Energy bands in intrinsic and extrinsic semiconductors; equilibrium carrier concentration; direct and indirect band-gap semiconductors; carrier transport (diffusion and drift current, mobility and resistivity, generation and recombination, Poisson and continuity equations); P-N junction, Zener diode, BJT, MOS capacitor, MOSFET, LED, photodiode and solar cell.

Key concepts and formulas

(q = 1.6 × 10⁻¹⁹ C; thermal voltage V_T = kT/q ≈ 25.9 mV at 300 K, often taken as 25 or 26 mV.)

Carrier concentration. Mass action law np = nᵢ² in equilibrium. In an n-type material with donor density N_D ≫ nᵢ: n ≈ N_D and p ≈ nᵢ²/N_D. For silicon at 300 K, nᵢ ≈ 1.5 × 10¹⁰ cm⁻³ (some books use 10¹⁰), and E_g ≈ 1.12 eV. Silicon and germanium are indirect band-gap; GaAs is direct, which is why it emits light efficiently.

Transport.

• Drift current density J = q(nμ_n + pμ_p)E; conductivity σ = q(nμ_n + pμ_p); resistivity ρ = 1/σ.
• Diffusion current J_n = qD_n dn/dx.
• Einstein relation: D/μ = V_T.
• Diffusion length L = √(Dτ).

P-N junction.

• Built-in potential V₀ = V_T ln(N_AN_D/nᵢ²).
• Depletion width W = √(2ε(V₀ + V_R)(1/N_A + 1/N_D)/q); the depletion region extends mostly into the lightly doped side. Junction capacitance C_j = εA/W ∝ (V₀ + V_R)^−1/2 for an abrupt junction.
• Diode equation I = I_S(e^V/ηV_T − 1). A 60 mV (at η = 1) increase in forward voltage multiplies the current by about 10. Dynamic resistance r_d = ηV_T/I.
• Zener breakdown (heavy doping, below about 5 V) has a negative temperature coefficient; avalanche breakdown (above about 6 V) a positive one.

BJT. Active region: emitter-base forward biased, collector-base reverse biased. I_C = βI_B, α = β/(1 + β). Early effect: output resistance r_o = V_A/I_C.

MOS capacitor and MOSFET.

• Threshold voltage V_T marks strong inversion. C_ox = ε_ox/t_ox.
• n-MOSFET in saturation (V_DS ≥ V_GS − V_T): I_D = ½μ_nC_ox(W/L)(V_GS − V_T)²; in the linear region: I_D = μ_nC_ox(W/L)[(V_GS − V_T)V_DS − V_DS²/2].
• Transconductance g_m = 2I_D/(V_GS − V_T) = √(2μ_nC_ox(W/L)I_D).

Optoelectronics. An LED emits at wavelength λ = hc/E_g, i.e. λ (μm) ≈ 1.24/E_g (eV). A photodiode works in reverse bias; a solar cell in the fourth quadrant (forward voltage, reverse current), with fill factor = P_max/(V_ocI_sc).

Worked example

Silicon is doped with N_D = 10¹⁶ cm⁻³ (nᵢ = 1.5 × 10¹⁰ cm⁻³, μ_n = 1350 cm²/Vs). Find n, p and the resistivity. Find the built-in potential of a junction with N_A = 10¹⁷ cm⁻³ and N_D = 10¹⁶ cm⁻³ at V_T = 25.9 mV, and the saturation current of a MOSFET with μ_nC_ox = 200 μA/V², W/L = 10, V_GS = 1.5 V and V_T = 0.5 V.

import math

q, ni, ND, mun = 1.6e-19, 1.5e10, 1e16, 1350
n, p = ND, ni ** 2 / ND
rho = 1 / (q * n * mun)
print(f"n = {n:.1e} cm^-3, p = {p:.3e} cm^-3, resistivity = {rho:.4f} ohm cm")
VT, NA = 0.0259, 1e17
print(f"built-in potential = {VT * math.log(NA * ND / ni ** 2):.4f} V")
k, WL, VGS, Vth = 200e-6, 10, 1.5, 0.5
ID = 0.5 * k * WL * (VGS - Vth) ** 2
print(f"ID(sat) = {ID * 1000:.3f} mA, gm = {2 * ID / (VGS - Vth) * 1000:.3f} mA/V")
print(f"LED with Eg = 1.42 eV emits at {1.24 / 1.42:.4f} um")

n = 1.0e+16 cm^-3, p = 2.250e+04 cm^-3, resistivity = 0.4630 ohm cm
built-in potential = 0.7543 V
ID(sat) = 1.000 mA, gm = 2.000 mA/V
LED with Eg = 1.42 eV emits at 0.8732 um

Practice questions